Meeting Rooms II: Heap Approach

Alternative: use a min-heap of ending times.

function minMeetingRooms(intervals)
    sort intervals by start time
    heap := min-heap of end times

    for interval in intervals
        if heap not empty and heap.peek() <= interval.start then
            // Reuse a room
            heap.pop()
        heap.push(interval.end)

    return heap.size()

The heap tracks rooms in use. When a meeting starts:

• If any room is free (earliest end time $\leq$ current start), reuse it
• Otherwise, add a new room

The heap size at the end is the maximum rooms needed.

Time: $O(n \log n)$. Space: $O(n)$.